在Java中从一个基数转换为另一个基数

Question

现在,我正在尝试找到一种方法,在Java中将数字从一个基数转换为另一个数字,给出一个数字,数字所在的基数,以及转换为的基数.

public static void BaseConversion(String number, int base1, int base2){
    //convert the number from one base to another
}

我找到了JavaScript的解决方案,我想知道是否可以在Java中做类似的事情:

function convertFromBaseToBase(str, fromBase, toBase){
    var num = parseInt(str, fromBase); //convert from one base to another
    return num.toString(toBase);
}

Answer 1

你可以做到

return Integer.toString(Integer.parseInt(number, base1), base2);

所以使用您的函数签名,在Java中:

public String convertFromBaseToBase(String str, int fromBase, int toBase) {
    return Integer.toString(Integer.parseInt(str, fromBase), toBase);
}

Answer 2

public class BaseToBaseConv {

static String baseToBase(String num, int base1, int base2) {
    int no = convFrmBaseToDeci(num, base1);
    return convFrmDecToBase(no, base2);
}

static String convFrmDecToBase(int num, int base) {

    String res = "";
    int rem;
    // Convert input number is given base by repeatedly
    // dividing it by base and taking remainder
    while (num > 0) {
        rem = num % base;
        if (base == 16) {
            if (rem == 10)
                res += 'A';
            else if (rem == 11)
                res += 'B';
            else if (rem == 12)
                res += 'C';
            else if (rem == 13)
                res += 'D';
            else if (rem == 14)
                res += 'E';
            else if (rem == 15)
                res += 'F';
            else
                res += rem;
        } else
            res += rem;

        num /= base;
    }
    // Reverse the result
    return new StringBuffer(res).reverse().toString();
}

static int convFrmBaseToDeci(String num, int base) {

    if (base < 2 || (base > 10 && base != 16))
        return -1;

    int val = 0;
    int power = 1;

    for (int i = num.length() - 1; i >= 0; i--) {
        int digit = digitToVal(num.charAt(i));

        if (digit < 0 || digit >= base)
            return -1;

        // Decimal equivalent is str[len-1]*1 +
        // str[len-1]*base + str[len-1]*(base^2) + ...
        val += digit * power;
        power = power * base;
    }

    return val;
}

static int digitToVal(char c) {
    if (c >= '0' && c <= '9')
        return (int) c - '0';
    else
        return (int) c - 'A' + 10;
}

public static void main(String [] args) {
    System.out.println(baseToBase("12345", 10, 2));
    System.out.println(baseToBase("11000000111001", 2, 10));
    System.out.println(baseToBase("ABC11", 16, 2));
    System.out.println(baseToBase("10101011110000010001", 2, 16));
    System.out.println(baseToBase("12322", 8, 16));
}
}

Answer 3

的两个参数的版本Integer.parseInt或Long.parseLong将做到这一点，如果你可以肯定有问题的数量范围内int或long分别。如果您不能保证这一点，请使用java.math.BigInteger：

BigInteger bi = new BigInteger(number, base1);
return bi.toString(base2);

这可以处理任意大的整数，例如

System.out.println(
  new BigInteger("12345678901234567890123456789", 10).toString(16));
// prints 27e41b3246bec9b16e398115 - too big to represent as a long

Answer 4

我相信这会起作用：

long x = 10;
int baseToConvertTo = 9;
System.out.println(Long.toString(x, baseToConvertTo));

输出： 11